Logo

AskSia

Plus

(c) Consider the random walk process Yt=Yt−1+ϵtY_{t}=Y_{t-1}+\epsilon_{t} where $\epsil...
May 30, 2024
Sure, let's break down the problem step by step.
Solution
a
Given the random walk process Yt=Yt−1+ϵtY_{t}=Y_{t-1}+\epsilon_{t}, where ϵt\epsilon_{t} is a zero mean white noise process with variance σϵ2\sigma_{\epsilon}^{2}, we need to find the variance of the sample mean Yˉn=∑t=1nYt/n\bar{Y}_{n}=\sum_{t=1}^{n} Y_{t} / n
b
The variance of the sample mean Yˉn\bar{Y}_{n} can be expressed as V[Yˉn]=1n2V[∑t=1nYt]V\left[\bar{Y}_{n}\right] = \frac{1}{n^2} V\left[\sum_{t=1}^{n} Y_{t}\right]
c
Since YtY_{t} is a random walk, Yt=Y0+∑i=1tϵiY_{t} = Y_{0} + \sum_{i=1}^{t} \epsilon_{i}. Therefore, V[Yt]=V[Y0+∑i=1tϵi]=∑i=1tV[ϵi]=tσϵ2V\left[Y_{t}\right] = V\left[Y_{0} + \sum_{i=1}^{t} \epsilon_{i}\right] = \sum_{i=1}^{t} V\left[\epsilon_{i}\right] = t \sigma_{\epsilon}^{2}
d
Now, V[∑t=1nYt]=∑t=1nV[Yt]=∑t=1ntσϵ2=σϵ2∑t=1nt=σϵ2n(n+1)2V\left[\sum_{t=1}^{n} Y_{t}\right] = \sum_{t=1}^{n} V\left[Y_{t}\right] = \sum_{t=1}^{n} t \sigma_{\epsilon}^{2} = \sigma_{\epsilon}^{2} \sum_{t=1}^{n} t = \sigma_{\epsilon}^{2} \frac{n(n+1)}{2}
e
Therefore, V[Yˉn]=1n2⋅σϵ2⋅n(n+1)2=σϵ2(n+1)2nV\left[\bar{Y}_{n}\right] = \frac{1}{n^2} \cdot \sigma_{\epsilon}^{2} \cdot \frac{n(n+1)}{2} = \frac{\sigma_{\epsilon}^{2} (n+1)}{2n}
f
As n→∞n \rightarrow \infty, V[Yˉn]→σϵ22V\left[\bar{Y}_{n}\right] \rightarrow \frac{\sigma_{\epsilon}^{2}}{2}. This indicates that the variance of the sample mean does not go to zero as nn increases, implying that the sample mean Yˉn\bar{Y}_{n} is not a reliable estimator of E[Yt]E\left[Y_{t}\right] for a random walk process
Answer
The variance of the sample mean Yˉn\bar{Y}_{n} is σϵ2(n+1)2n\frac{\sigma_{\epsilon}^{2} (n+1)}{2n}, and as n→∞n \rightarrow \infty, it approaches σϵ22\frac{\sigma_{\epsilon}^{2}}{2}. This means that the sample mean is not a reliable estimator of the expected value of YtY_{t} in a random walk process.
Key Concept
Variance of the sample mean in a random walk process
Explanation
The variance of the sample mean Yˉn\bar{Y}_{n} does not go to zero as the sample size nn increases, indicating that the sample mean is not a reliable estimator of the expected value of YtY_{t} in a random walk process.
Solution
a
Given the process Zt=ϵt−θϵt−12Z_{t}=\epsilon_{t}-\theta \epsilon_{t-12}, where ϵt\epsilon_{t} is a zero mean white noise process with variance σϵ2\sigma_{\epsilon}^{2}, we need to derive the autocorrelation function of ZtZ_{t}
b
The autocorrelation function ρk\rho_k of ZtZ_t is defined as ρk=γkγ0\rho_k = \frac{\gamma_k}{\gamma_0}, where γk\gamma_k is the autocovariance function at lag kk and γ0\gamma_0 is the variance of ZtZ_t
c
First, calculate the variance γ0\gamma_0 of ZtZ_t: γ0=E[Zt2]=E[(ϵt−θϵt−12)2]=E[ϵt2]+θ2E[ϵt−122]=σϵ2+θ2σϵ2=(1+θ2)σϵ2 \gamma_0 = E[Z_t^2] = E[(\epsilon_t - \theta \epsilon_{t-12})^2] = E[\epsilon_t^2] + \theta^2 E[\epsilon_{t-12}^2] = \sigma_{\epsilon}^2 + \theta^2 \sigma_{\epsilon}^2 = (1 + \theta^2) \sigma_{\epsilon}^2
d
Next, calculate the autocovariance γk\gamma_k for k≠0k \neq 0: γk=E[ZtZt−k]=E[(ϵt−θϵt−12)(ϵt−k−θϵt−12−k)] \gamma_k = E[Z_t Z_{t-k}] = E[(\epsilon_t - \theta \epsilon_{t-12})(\epsilon_{t-k} - \theta \epsilon_{t-12-k})] For k=12k = 12: γ12=E[(ϵt−θϵt−12)(ϵt−12−θϵt−24)]=−θσϵ2 \gamma_{12} = E[(\epsilon_t - \theta \epsilon_{t-12})(\epsilon_{t-12} - \theta \epsilon_{t-24})] = -\theta \sigma_{\epsilon}^2 For k=0k = 0: γ0=(1+θ2)σϵ2 \gamma_0 = (1 + \theta^2) \sigma_{\epsilon}^2 For k≠0,12k \neq 0, 12: γk=0 \gamma_k = 0
e
Finally, the autocorrelation function ρk\rho_k is: ρk=γkγ0 \rho_k = \frac{\gamma_k}{\gamma_0} For k=12k = 12: ρ12=−θσϵ2(1+θ2)σϵ2=−θ1+θ2 \rho_{12} = \frac{-\theta \sigma_{\epsilon}^2}{(1 + \theta^2) \sigma_{\epsilon}^2} = \frac{-\theta}{1 + \theta^2} For k≠0,12k \neq 0, 12: ρk=0 \rho_k = 0
Answer
The autocorrelation function of ZtZ_t is ρ12=−θ1+θ2\rho_{12} = \frac{-\theta}{1 + \theta^2} for k=12k = 12 and ρk=0\rho_k = 0 for k≠0,12k \neq 0, 12.
Key Concept
Autocorrelation function of a time series process
Explanation
The autocorrelation function measures the correlation between values of the process at different times. For the given process ZtZ_t, the autocorrelation is non-zero only at lag 12, indicating a specific periodic relationship.
Solution
a
Define the process: The given process is Xt=ϕXt−4+ϵtX_{t}=\phi X_{t-4}+\epsilon_{t}, where |\phi|<1 and ϵt\epsilon_{t} is a zero mean white noise process with variance σϵ2\sigma_{\epsilon}^{2}
b
Calculate the mean: Since ϵt\epsilon_{t} is a zero mean white noise process, the mean of XtX_{t} is also zero
c
Calculate the variance: The variance of XtX_{t} can be derived using the fact that ϵt\epsilon_{t} is white noise with variance σϵ2\sigma_{\epsilon}^{2}. The variance of XtX_{t} is given by σX2=σϵ21−ϕ2\sigma_{X}^{2} = \frac{\sigma_{\epsilon}^{2}}{1-\phi^2}
d
Define the autocovariance function: The autocovariance function γk\gamma_k for lag kk is defined as γk=E[(Xt−μ)(Xt−k−μ)]\gamma_k = \mathbb{E}[(X_t - \mu)(X_{t-k} - \mu)]. Since μ=0\mu = 0, this simplifies to γk=E[XtXt−k]\gamma_k = \mathbb{E}[X_t X_{t-k}]
e
Calculate the autocovariance for lag 0: For k=0k=0, γ0=E[Xt2]=σX2=σϵ21−ϕ2\gamma_0 = \mathbb{E}[X_t^2] = \sigma_{X}^{2} = \frac{\sigma_{\epsilon}^{2}}{1-\phi^2}
f
Calculate the autocovariance for lag 4: For k=4k=4, γ4=E[XtXt−4]=ϕE[Xt−42]=ϕσX2=ϕσϵ21−ϕ2\gamma_4 = \mathbb{E}[X_t X_{t-4}] = \phi \mathbb{E}[X_{t-4}^2] = \phi \sigma_{X}^{2} = \phi \frac{\sigma_{\epsilon}^{2}}{1-\phi^2}
g
Calculate the autocovariance for other lags: For k≠0,4k \neq 0, 4, γk=0\gamma_k = 0 because XtX_t and Xt−kX_{t-k} are uncorrelated for these lags
h
Define the autocorrelation function: The autocorrelation function ρk\rho_k is given by ρk=γkγ0\rho_k = \frac{\gamma_k}{\gamma_0}
i
Calculate the autocorrelation for lag 0: For k=0k=0, ρ0=γ0γ0=1\rho_0 = \frac{\gamma_0}{\gamma_0} = 1
j
Calculate the autocorrelation for lag 4: For k=4k=4, ρ4=γ4γ0=ϕ\rho_4 = \frac{\gamma_4}{\gamma_0} = \phi
k
Calculate the autocorrelation for other lags: For k≠0,4k \neq 0, 4, ρk=0\rho_k = 0
Answer
The autocorrelation function of the process Xt=ϕXt−4+ϵtX_{t}=\phi X_{t-4}+\epsilon_{t} is given by: - ρ0=1\rho_0 = 1 - ρ4=ϕ\rho_4 = \phi - ρk=0\rho_k = 0 for k≠0,4k \neq 0, 4
Key Concept
Autocorrelation Function
Explanation
The autocorrelation function measures the correlation between values of the process at different times. For the given process, the autocorrelation is non-zero only at lags 0 and 4, reflecting the structure of the process.
Solution
a
Expectation of XtX_t: Since XtX_t is a random walk process starting at X0=0X_0 = 0 and wtw_t is an IID Bernoulli process with equal probability of +a+a and −a-a, the expected value of XtX_t is E[Xt]=E[Xt−1+wt]=E[Xt−1]+E[wt]E[X_t] = E[X_{t-1} + w_t] = E[X_{t-1}] + E[w_t]. Given E[wt]=0E[w_t] = 0, we have E[Xt]=0E[X_t] = 0
b
Variance of XtX_t: The variance of XtX_t can be calculated as V[Xt]=V[Xt−1+wt]=V[Xt−1]+V[wt]V[X_t] = V[X_{t-1} + w_t] = V[X_{t-1}] + V[w_t] since Xt−1X_{t-1} and wtw_t are independent. Given V[wt]=a2V[w_t] = a^2, we have V[Xt]=t⋅a2V[X_t] = t \cdot a^2
Answer
(a) E[Xt]=0,V[Xt]=ta2E[X_t] = 0, V[X_t] = t a^2
Key Concept
Random Walk Process
Explanation
In a random walk process with IID steps, the expectation remains zero due to the equal probability of positive and negative steps, while the variance accumulates linearly with time.
Solution
a
Definition of the MA(2) process: The given process is Zt=εt+θ1εt−1+θ2εt−2Z_{t} = \varepsilon_{t} + \theta_{1} \varepsilon_{t-1} + \theta_{2} \varepsilon_{t-2} where εt∼IID⁡(0,1)\varepsilon_{t} \sim \operatorname{IID}(0,1) and θ1,θ2\theta_{1}, \theta_{2} are constants
b
Sample Mean: The sample mean is 14(Z1+Z2+Z3+Z4)\frac{1}{4}(Z_{1} + Z_{2} + Z_{3} + Z_{4})
c
Variance of the Sample Mean: The variance of the sample mean is given by Var⁡(14(Z1+Z2+Z3+Z4))\operatorname{Var}\left(\frac{1}{4}(Z_{1} + Z_{2} + Z_{3} + Z_{4})\right)
d
Calculation of Variance: Using the properties of variance and the given MA(2) process, we calculate the variance as follows: Var⁡(14(Z1+Z2+Z3+Z4))=116(Var⁡(Z1)+Var⁡(Z2)+Var⁡(Z3)+Var⁡(Z4)+2Cov⁡(Z1,Z2)+2Cov⁡(Z1,Z3)+2Cov⁡(Z1,Z4)+2Cov⁡(Z2,Z3)+2Cov⁡(Z2,Z4)+2Cov⁡(Z3,Z4)) \operatorname{Var}\left(\frac{1}{4}(Z_{1} + Z_{2} + Z_{3} + Z_{4})\right) = \frac{1}{16} \left( \operatorname{Var}(Z_{1}) + \operatorname{Var}(Z_{2}) + \operatorname{Var}(Z_{3}) + \operatorname{Var}(Z_{4}) + 2 \operatorname{Cov}(Z_{1}, Z_{2}) + 2 \operatorname{Cov}(Z_{1}, Z_{3}) + 2 \operatorname{Cov}(Z_{1}, Z_{4}) + 2 \operatorname{Cov}(Z_{2}, Z_{3}) + 2 \operatorname{Cov}(Z_{2}, Z_{4}) + 2 \operatorname{Cov}(Z_{3}, Z_{4}) \right) Given the MA(2) process, we know: Var⁡(Zt)=1+θ12+θ22 \operatorname{Var}(Z_{t}) = 1 + \theta_{1}^{2} + \theta_{2}^{2} Cov⁡(Zt,Zt−1)=θ1+θ1θ2 \operatorname{Cov}(Z_{t}, Z_{t-1}) = \theta_{1} + \theta_{1}\theta_{2} Cov⁡(Zt,Zt−2)=θ2 \operatorname{Cov}(Z_{t}, Z_{t-2}) = \theta_{2} Cov⁡(Zt,Zt−3)=0 \operatorname{Cov}(Z_{t}, Z_{t-3}) = 0 Substituting these into the variance formula, we get: Var⁡(14(Z1+Z2+Z3+Z4))=116(4(1+θ12+θ22)+6θ1+4θ1θ2+2θ2) \operatorname{Var}\left(\frac{1}{4}(Z_{1} + Z_{2} + Z_{3} + Z_{4})\right) = \frac{1}{16} \left( 4(1 + \theta_{1}^{2} + \theta_{2}^{2}) + 6\theta_{1} + 4\theta_{1}\theta_{2} + 2\theta_{2} \right) Simplifying, we get: Var⁡(14(Z1+Z2+Z3+Z4))=(1+θ12+θ22)4+(6θ1(1+θ2)+4θ2)16 \operatorname{Var}\left(\frac{1}{4}(Z_{1} + Z_{2} + Z_{3} + Z_{4})\right) = \frac{(1 + \theta_{1}^{2} + \theta_{2}^{2})}{4} + \frac{(6\theta_{1}(1 + \theta_{2}) + 4\theta_{2})}{16}
Answer
(c) (1+θ12+θ22)4+(6θ1(1+θ2)+4θ2)16\frac{(1 + \theta_{1}^{2} + \theta_{2}^{2})}{4} + \frac{(6\theta_{1}(1 + \theta_{2}) + 4\theta_{2})}{16}
Key Concept
Variance of the sample mean in an MA(2) process
Explanation
The variance of the sample mean for the given MA(2) process is derived by considering the variances and covariances of the individual terms in the process. The correct expression matches option (c).
Solution
a
Definition of AR(1) Process: An AR(1) process is defined as Yt=ϕYt−1+ϵtY_t = \phi Y_{t-1} + \epsilon_t, where ϵt\epsilon_t is white noise. In this case, Yt=2.5+0.7Yt−1+ϵtY_t = 2.5 + 0.7 Y_{t-1} + \epsilon_t
b
Causality: The model is causal if |\phi| < 1. Here, ϕ=0.7\phi = 0.7, so the model is causal
c
First Order Autocorrelation: The first order autocorrelation ρ1\rho_1 is equal to ϕ\phi. Therefore, ρ1=0.7\rho_1 = 0.7
d
Second Order Autocovariance: The second order autocovariance γ2\gamma_2 can be calculated using γ2=ϕ2γ0\gamma_2 = \phi^2 \gamma_0. Given γ0=σϵ21−ϕ2\gamma_0 = \frac{\sigma_\epsilon^2}{1-\phi^2}, we find γ0=91−0.49=17.65\gamma_0 = \frac{9}{1-0.49} = 17.65. Thus, γ2=0.72×17.65=8.65\gamma_2 = 0.7^2 \times 17.65 = 8.65
e
Variance of YtY_t: The variance V[Yt]V[Y_t] is given by γ0=σϵ21−ϕ2\gamma_0 = \frac{\sigma_\epsilon^2}{1-\phi^2}. Therefore, V[Yt]=17.65V[Y_t] = 17.65, not 15
Answer
The incorrect statement is (d) The variance of YtY_t is V[Yt]=15V[Y_t] = 15.
Key Concept
Variance Calculation in AR(1) Process
Explanation
The variance of an AR(1) process is calculated using γ0=σϵ21−ϕ2\gamma_0 = \frac{\sigma_\epsilon^2}{1-\phi^2}. In this case, the variance is 17.6517.65, not 1515.
Solution
a
Causality: The given process rt=0.9+0.5rt−1+ϵtr_{t}=0.9+0.5 r_{t-1}+\epsilon_{t} is causal because it can be expressed in terms of past values of rtr_t and the white noise error term ϵt\epsilon_t
b
Autocorrelation Coefficient: The autocorrelation coefficient of order 3, ρ3\rho_{3}, is calculated using the formula for AR(1) processes. For an AR(1) process, ρk=ϕk\rho_k = \phi^k. Here, ϕ=0.5\phi = 0.5, so ρ3=(0.5)3=0.125\rho_{3} = (0.5)^3 = 0.125
c
Partial Derivative at t+3t+3: The partial derivative ∂rt+3∂ϵt\frac{\partial r_{t+3}}{\partial \epsilon_{t}} is calculated by considering the impact of ϵt\epsilon_t on rt+3r_{t+3}. For an AR(1) process, this is given by ϕ3=(0.5)3=0.125\phi^3 = (0.5)^3 = 0.125
d
Partial Derivative at t+100t+100: The partial derivative ∂rt+100∂ϵt\frac{\partial r_{t+100}}{\partial \epsilon_{t}} is calculated similarly. For an AR(1) process, this is given by ϕ100=(0.5)100\phi^{100} = (0.5)^{100}, which is a very small number, not 0.150
Answer
The incorrect statement is (d) ∂rt+100∂ϵt=0.150\frac{\partial r_{t+100}}{\partial \epsilon_{t}}=0.150.
Key Concept
Partial derivatives in AR(1) processes decrease exponentially with the lag.
Explanation
In an AR(1) process, the impact of a shock ϵt\epsilon_t on future values rt+kr_{t+k} diminishes exponentially as kk increases. Therefore, ∂rt+100∂ϵt\frac{\partial r_{t+100}}{\partial \epsilon_{t}} should be a very small number, not 0.150.
Solution
a
Given the AR(1) process Xt=ϕXt−1+ϵtX_t = \phi X_{t-1} + \epsilon_t where ϵt\epsilon_t is a zero mean white noise process with variance σϵ2\sigma_{\epsilon}^2 and |\phi| < 1, we need to analyze the first difference Zt=Xt−Xt−1Z_t = X_t - X_t-1
b
The variance of XtX_t is given by V[Xt]=σϵ21−ϕ2V[X_t] = \frac{\sigma_{\epsilon}^2}{1 - \phi^2}
c
The variance of ZtZ_t is given by V[Zt]=V[Xt−Xt−1]=V[Xt]+V[Xt−1]−2Cov(Xt,Xt−1)V[Z_t] = V[X_t - X_{t-1}] = V[X_t] + V[X_{t-1}] - 2 \text{Cov}(X_t, X_{t-1})
d
Since XtX_t is an AR(1) process, Cov(Xt,Xt−1)=ϕV[Xt−1]=ϕσϵ21−ϕ2\text{Cov}(X_t, X_{t-1}) = \phi V[X_{t-1}] = \phi \frac{\sigma_{\epsilon}^2}{1 - \phi^2}
e
Therefore, V[Zt]=2σϵ21−ϕ2−2ϕσϵ21−ϕ2=2(1−ϕ)σϵ21−ϕ2V[Z_t] = 2 \frac{\sigma_{\epsilon}^2}{1 - \phi^2} - 2 \phi \frac{\sigma_{\epsilon}^2}{1 - \phi^2} = 2(1 - \phi) \frac{\sigma_{\epsilon}^2}{1 - \phi^2}
f
Comparing the variances, V[Zt]=2(1−ϕ)σϵ21−ϕ2V[Z_t] = 2(1 - \phi) \frac{\sigma_{\epsilon}^2}{1 - \phi^2} and V[Xt]=σϵ21−ϕ2V[X_t] = \frac{\sigma_{\epsilon}^2}{1 - \phi^2}
g
For ϕ=0\phi = 0, V[Zt]=2σϵ2V[Z_t] = 2 \sigma_{\epsilon}^2 and V[Xt]=σϵ2V[X_t] = \sigma_{\epsilon}^2, so V[Z_t] > V[X_t]
h
For ϕ=12\phi = \frac{1}{2}, V[Zt]=2(1−12)σϵ21−(12)2=43σϵ2V[Z_t] = 2(1 - \frac{1}{2}) \frac{\sigma_{\epsilon}^2}{1 - (\frac{1}{2})^2} = \frac{4}{3} \sigma_{\epsilon}^2 and V[Xt]=43σϵ2V[X_t] = \frac{4}{3} \sigma_{\epsilon}^2, so V[Zt]=V[Xt]V[Z_t] = V[X_t]
i
Therefore, the incorrect statement is (b) The variance of ZtZ_t is always larger than the variance of XtX_t
Answer
The incorrect statement is (b) The variance of ZtZ_t is always larger than the variance of XtX_t.
Key Concept
Variance comparison in AR(1) processes
Explanation
The variance of the first difference ZtZ_t is not always larger than the variance of XtX_t. It depends on the value of ϕ\phi. For ϕ=12\phi = \frac{1}{2}, the variances are equal.
Solution
a
Given the process Xt=Xt−1+ϵtX_t = X_{t-1} + \epsilon_t, where ϵt\epsilon_t is a zero mean white noise process with variance σϵ2\sigma_{\epsilon}^2, we need to determine the correct autocorrelation function ρ(t,s)\rho(t, s)
b
The process XtX_t is a random walk, which implies that the increments are independent. Therefore, the autocorrelation function ρ(t,s)\rho(t, s) for a random walk is 1 for all tt and ss
c
Thus, the correct statement is (a) ρ(t,s)=1\rho(t, s) = 1 for all tt and ss
Answer
(a) ρ(t,s)=1\rho(t, s) = 1 for all tt and ss
Key Concept
Autocorrelation function of a random walk
Explanation
In a random walk, the increments are independent, leading to an autocorrelation function of 1 for all time periods.
How does the variance of the sample mean V[Yˉn]V[\bar{Y}_{n}] change in relation to the sample size nn as n→∞n \rightarrow \infty for a random walk process YtY_{t}?
Solution
a
Definition of Random Walk: A random walk process YtY_t is defined as Yt=Yt−1+ϵtY_t = Y_{t-1} + \epsilon_t, where ϵt\epsilon_t is a white noise error term with mean zero and variance σ2\sigma^2
b
Sample Mean: The sample mean Yˉn\bar{Y}_n for a random walk process is given by Yˉn=1n∑t=1nYt\bar{Y}_n = \frac{1}{n} \sum_{t=1}^n Y_t
c
Variance of Sample Mean: The variance of the sample mean V[Yˉn]V[\bar{Y}_n] can be expressed as V[Yˉn]=1n2∑t=1n∑s=1nCov(Yt,Ys)V[\bar{Y}_n] = \frac{1}{n^2} \sum_{t=1}^n \sum_{s=1}^n \text{Cov}(Y_t, Y_s)
d
Covariance in Random Walk: For a random walk, Cov(Yt,Ys)=σ2min⁡(t,s)\text{Cov}(Y_t, Y_s) = \sigma^2 \min(t, s)
e
Simplifying Variance: Substituting the covariance into the variance expression, we get V[Yˉn]=1n2∑t=1n∑s=1nσ2min⁡(t,s)V[\bar{Y}_n] = \frac{1}{n^2} \sum_{t=1}^n \sum_{s=1}^n \sigma^2 \min(t, s)
f
Asymptotic Behavior: As n→∞n \rightarrow \infty, the double sum ∑t=1n∑s=1nmin⁡(t,s)\sum_{t=1}^n \sum_{s=1}^n \min(t, s) is approximately n33\frac{n^3}{3}, leading to V[Yˉn]≈σ2n33n2=σ2n3V[\bar{Y}_n] \approx \frac{\sigma^2 n^3}{3n^2} = \frac{\sigma^2 n}{3}
g
Conclusion: Therefore, V[Yˉn]V[\bar{Y}_n] increases linearly with nn as n→∞n \rightarrow \infty
Answer
The variance of the sample mean V[Yˉn]V[\bar{Y}_n] increases linearly with the sample size nn as n→∞n \rightarrow \infty for a random walk process YtY_t.
Key Concept
Variance of Sample Mean in Random Walk
Explanation
For a random walk process, the variance of the sample mean V[Yˉn]V[\bar{Y}_n] increases linearly with the sample size nn as nn approaches infinity. This is due to the cumulative nature of the random walk, where the variance of the sum of the terms grows with the number of terms.
Solution
a
Given the random walk process Yt=Yt−1+ϵtY_{t}=Y_{t-1}+\epsilon_{t}, where ϵt\epsilon_{t} is a zero mean white noise process with variance σϵ2\sigma_{\epsilon}^{2}, we need to find the variance of the sample mean Yˉn=∑t=1nYt/n\bar{Y}_{n}=\sum_{t=1}^{n} Y_{t} / n
b
First, note that YtY_{t} can be expressed as Yt=Y0+∑i=1tϵiY_{t} = Y_{0} + \sum_{i=1}^{t} \epsilon_{i}
c
The sample mean Yˉn\bar{Y}_{n} is then Yˉn=1n∑t=1nYt=1n∑t=1n(Y0+∑i=1tϵi)\bar{Y}_{n} = \frac{1}{n} \sum_{t=1}^{n} Y_{t} = \frac{1}{n} \sum_{t=1}^{n} \left( Y_{0} + \sum_{i=1}^{t} \epsilon_{i} \right)
d
Simplifying, we get Yˉn=Y0+1n∑t=1n∑i=1tϵi\bar{Y}_{n} = Y_{0} + \frac{1}{n} \sum_{t=1}^{n} \sum_{i=1}^{t} \epsilon_{i}
e
The variance of Yˉn\bar{Y}_{n} is V[Yˉn]=V[Y0+1n∑t=1n∑i=1tϵi]V[\bar{Y}_{n}] = V\left[ Y_{0} + \frac{1}{n} \sum_{t=1}^{n} \sum_{i=1}^{t} \epsilon_{i} \right]. Since Y0Y_{0} is a constant, V[Y0]=0V[Y_{0}] = 0
f
Therefore, V[Yˉn]=V[1n∑t=1n∑i=1tϵi]=1n2V[∑t=1n∑i=1tϵi]V[\bar{Y}_{n}] = V\left[ \frac{1}{n} \sum_{t=1}^{n} \sum_{i=1}^{t} \epsilon_{i} \right] = \frac{1}{n^2} V\left[ \sum_{t=1}^{n} \sum_{i=1}^{t} \epsilon_{i} \right]
g
Since ϵt\epsilon_{t} are independent with variance σϵ2\sigma_{\epsilon}^{2}, V[∑t=1n∑i=1tϵi]=∑t=1n∑i=1tV[ϵi]=σϵ2∑t=1nt=σϵ2n(n+1)2V\left[ \sum_{t=1}^{n} \sum_{i=1}^{t} \epsilon_{i} \right] = \sum_{t=1}^{n} \sum_{i=1}^{t} V[\epsilon_{i}] = \sigma_{\epsilon}^{2} \sum_{t=1}^{n} t = \sigma_{\epsilon}^{2} \frac{n(n+1)}{2}
h
Thus, V[Yˉn]=1n2⋅σϵ2⋅n(n+1)2=σϵ2(n+1)2nV[\bar{Y}_{n}] = \frac{1}{n^2} \cdot \sigma_{\epsilon}^{2} \cdot \frac{n(n+1)}{2} = \frac{\sigma_{\epsilon}^{2} (n+1)}{2n}
Answer
σϵ2(n+1)2n\frac{\sigma_{\epsilon}^{2} (n+1)}{2n}
Key Concept
Variance of the sample mean in a random walk process
Explanation
The variance of the sample mean Yˉn\bar{Y}_{n} for a random walk process Yt=Yt−1+ϵtY_{t}=Y_{t-1}+\epsilon_{t}, where ϵt\epsilon_{t} is a zero mean white noise process with variance σϵ2\sigma_{\epsilon}^{2}, is derived by considering the sum of variances of the white noise terms and normalizing by the sample size.
© 2023 AskSia.AI all rights reserved